Factor the following expression: $ x^2 + 13xy + 40y^2 $
Answer: When we factor a polynomial of this form, we are basically reversing this process of multiplying linear expressions together: $ \begin{eqnarray} (x + ay)(x + by)&=&xx &+& xby + ayx &+& ayby \\ \\ &=& x^2 &+& {(a+b)}xy &+& {ab}y^2 \\ &\hphantom{=}& \hphantom{x^2} &\hphantom{+}& \hphantom{{13}xy} &\hphantom{+}& \hphantom{{40}y^2} \end{eqnarray} $ $ \begin{eqnarray} \hphantom{(x + ay)(x + by)}&\hphantom{=}&\hphantom{xx} &\hphantom{+}& \hphantom{xby + ayx} &\hphantom{+}&\hphantom{ayby} \\ &\hphantom{=}& \hphantom{x^2} &\hphantom{+}&\hphantom{{(a+b)}xy}&\hphantom{+}&\hphantom{{ab}y^2} \\ &=& x^2 &+& {13}xy &+& {40}y^2 \end{eqnarray} $ The coefficient on the $xy$ term is $13$ and the coefficient on the $y^2$ term is $40$ , so to reverse the steps above, we need to find two numbers that add up to $13$ and multiply to $40$ You can start by trying to guess which factors of $40$ add up to $13$ . In other words, you need to find the values for $a$ and $b$ that meet the following conditions: $ {a} + {b} = {13}$ $ {a} \times {b} = {40}$ If you're stuck, try listing out every single factor of $40$ and its opposite as $a$ in these equations, and see if it gives a value for $b$ that validates both conditions. For example, since $5$ is a factor of $40$ , try substituting $5$ for $a$ as well as $-5$ The two numbers $5$ and $8$ satisfy both conditions: $ {5} + {8} = {13} $ $ {5} \times {8} = {40} $ So we can factor the polynomial as $(x + 5y)(x + 8y)$.